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What_to_study_for_the_First_Mid - Topics to be mastered for...

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Topics to be mastered for the second mid-term exam in CHEM6C : Remember that for this exam (and for the final) only the commercial, laminated study guide is allowed. 1. Writing chemical equations for various kinds of reactions, e.g., write an equation showing the results of adding a basic anhydride to water, or write an equation showing the use of Cl 2 as an oxidizing agent, etc. Basic anhydride - N 2 O KO 2 BaO CaO ( group I and II with O) Group I and II reacting with water give basic solution (OH-) Na + H2O --> 2NaOH + H2 NaH + H2O ---> NaOH + H2 Na2O2 + H2O ---> NaOH + H2O2 Same goes to BaO and CaO (group II) BaO + H2O ---> Ba(OH)2 ... Cl2 reaction: oxidizing power Cl2 > Br2> I2 (see redution potential table) 2Fe + 3Cl2 ---> 2FeCl3 2Br- + Cl2 ----> Br2 + 2Cl- 2I- + Cl2 ---> I2 + Cl- Cl2 + H2O ----> HClO + HCl Cl2 + 2OH- ----> ClO- + Cl- + H2O Predicting reaction - Trend MgSi + H2O ----> SiH4 K3P + H2O ----> PH3 Li4C + H2O ----> CH4 +2 -1 +1 -1 +2 -1 +1 -2 Mg(OH)2 + HCl ---> MgCl2 + H2O way to think of it: determine ox. no of each atoms, -ve species should react with +ve species of another molecule, and vice versa. (Note that Mg charge is 2+ so you need 2 Cl- . Same goes to H and O) +3 -2 0 +2 -2 0 B2O3 + Mg ---> MgO + B Mg is group II element, likely to lose 2e- and become Mg2+ and form bond with -ve species, which is O). B accept s e- to become B. 0 0 +3 -1 2Al + 3Cl 2 ---> 2AlCl3 Al is group III element, likely to lose 3e- and become Al3+ (to complete odctet) and Cl2 is likely to form Cl- (to complete octet).
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2Al + O2 ---> Al 2 O 3 same idea as above. 2. Allotropic behavior of the main-group elements. group IV C allotrope graphite - sp2 hybridization - pi e- delocalized - electrically conductive - intermolecular force between layers is van der waal - lower boiling pt diamond - sp3 hybridation - no pi e- delocalized - not electracally conductive - intermolecular force is covalent bond (network) - strong - higher pt other allotrope graphene C60 CO2 vs SiO2 CO2 - double bond because atomic radius are similar between C and O, similar size in p orbitals, leading to effectively overlap. Form linear molecule. SiO2 - Si is larger than O, radius between nuclei are further away. P also has larger and more diffuse p orbitals, leading to less effectively overlap- then not likely to form double bond. Form single bond with 4 O and have tetrahedral shape instead. group V - N2 triple bonds same idea as CO2 - small radius- two nuclei close to each other- effectively overlap - P4 - single bond- not likely to form triple bond because atomic radius gets bigger, p orbitals overlap noneffectively. Likely to form single bond
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